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JamesRV

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We just started learning Conservation of Momentum, and I feel completely lost. This is the following question I was asked:

A 2.00-kg object moving at 6.00 m/s collides with a 4.00-kg object that is initially at rest. After the collision, the 2.00-kg object moves backward at 1.00 m/s. This is a one-dimensional collision, and no external forces are acting.

a) Calculate the velocity of the 4.00-kg object after the collision.

My attempt was to simply use the Momentum equation:

m

So:

v

Resulting in the velocity to be 3.5m/s.

I was just looking for someone to confirm that answer, because I am unsure if I did it right.

In question b) it asks:

Calculate the velocity of the centre of mass of the two objects (i) before the collision and (ii) after the collision.

What does that mean?

Thank you for helping.

## Homework Statement

A 2.00-kg object moving at 6.00 m/s collides with a 4.00-kg object that is initially at rest. After the collision, the 2.00-kg object moves backward at 1.00 m/s. This is a one-dimensional collision, and no external forces are acting.

a) Calculate the velocity of the 4.00-kg object after the collision.

My attempt was to simply use the Momentum equation:

## Homework Equations

& The attempt at a solution[/B]m

_{A1}v_{A1}+ m_{B1}v_{B1}= m_{A2}v_{A2}+ m_{B2}v_{B2}So:

v

_{B2}= (m_{A1}v_{A1}+ m_{B1}v_{B1}- m_{A2}v_{A2})/m_{B2}Resulting in the velocity to be 3.5m/s.

I was just looking for someone to confirm that answer, because I am unsure if I did it right.

In question b) it asks:

Calculate the velocity of the centre of mass of the two objects (i) before the collision and (ii) after the collision.

What does that mean?

Thank you for helping.

Last edited: